All categories

3 years ago

B. The role of the moon is greater than that of the sun in the occurrence of tides. ???

1 answer:

7 0

Our sun is 27 million times larger than our moon. Based on its mass, the sun's gravitational attraction to the Earth is more than 177 times greater than that of the moon to the Earth. If tidal forces were based solely on comparative masses, the sun should have a tide-generating force that is 27 million times greater than that of the moon. However, the sun is 390 times further from the Earth than is the moon. Thus, its tide-generating force is reduced by 3903, or about 59 million times less than the moon. Because of these conditions, the sun’s tide-generating force is about half that of the moon.

You might be interested in

1. Calculate the average velocity of the following trip. You walk to Pershing Square 58

Gre4nikov [31]

Explanation:

Velocity = displacement / time

v = √((58 m)² + (135 m)²) / (12 min × 60 s/min)

v = 0.20 m/s

7 0

3 years ago

Glycerin is poured into an open U-shaped tube until the height in both sides is 20cm. Ethyl alcohol is then poured into one arm

Liula [17]

Answer:

7.5 cm

Explanation:

In the figure we can see a sketch of the problem. We know that at the bottom of the U-shaped tube the pressure is equal in both branches. Defining $\rho_A:$ Ethyl alcohol density and $\rho_G:$ Glycerin density , we can write:

$\rho_A\times g \times h_1 + \rho_G \times g \times h_2 = \rho_G \times g \times h_3$

Simplifying:

$\rho_A\times h_1 = \rho_G \times (h_3 - h_2) (1)$

On the other hand:

$h_1 + h_2 = \Delta h + h_3$

Rearranging:

$h_1 - \Delta h = h_3 - h_2 (2)$

Replacing (2) in (1):

$\rho_A\times h_1 = \rho_G \times (h_1 - \Delta h)$

Rearranging:

$\frac{h_1 \times (\rho_A - \rho_G)}{- \rho_G} = \Delta h$

Data:

$h_1 = 20 cm; \rho_A = 0.789 \frac{g}{cm^3}; \rho_G = 1.26 \frac{g}{cm^3}$

$\frac{20 cm \times (0.789 - 1.26) \frac{g}{cm^3}}{- 1.26\frac{g}{cm^3}} = \Delta h$

$7.5 cm = \Delta h$

7 0

3 years ago

The Displacement is 5m. We found that using the

Llana [10]

Answer:

A vector can be written as:

(R, θ)

Where R is the magnitude, in this case, we know that the magnitude of the displacement is 5m

Then:

R = 5m

and θ defines the direction, it's an angle measured from the positive x-axis.

(In the image, θ would be the angle located at the point A)

Now, if you look at the image, you can see a triangle rectangle.

Where the adjacent cathetus has a length of 4,

the opposite cathetus has a length of 3 units

the hypotenuse has a length of 5 units.

So we can use any trigonometric rule to find the value of θ, like:

sin(θ) = (opposite cathetus)/hypotenuse

Then:

sin(θ) = 3m/5m

Now we can use the inverse sin function, Asin(x), in both sides

Asin( sin(θ)) = θ = Asin( 3/5) = 36.87°

then the vector is:

(5m, 36.87°)

Now, if we define the positive y-axis as the North, and the positive x-axis as the East.

This vector would point at 36.87° North of East.

(or almost Northeast)

7 0

3 years ago

How can a 1kg ball have more kinetic energy than a 100kg ball? Explain both using words and by providing a numerical example

MariettaO [177]

1 kg ball can have more kinetic energy than a 100 kg ball as increase in velocity is having greater impact on K.E than increase in mass.

<u>Explanation</u>:

We know kinetic energy can be judged or calculated by two parameters only which is mass and velocity. As kinetic energy is directly proportional to the $(velocity)^2$ and increase in velocity leads to greater effect on translational Kinetic Energy. Here formula of Kinetic Energy suggests that doubling the mass will double its K.E but doubling velocity will quadruple its velocity:

$\text { Kinetic Energy }=\frac{1}{2} m v^{2}$

Better understood from numerical example as given:

If a man A having weight 50 kg run with speed 5 m/s and another man B having 100 kg weight run with 2.5 m / s. Which man will have more K.E?

This can be solved as follows:

$\text { Kinetic Energy of } \mathrm{A}=\frac{1}{2} 50 \times 5^{2}=625 \mathrm{J}$

$\text { Kinetic Energy bf } \mathrm{B}=\frac{1}{2} 100 \times 2.5^{2}=312.5 \mathrm{J}$

It shows that man A will have more K.E.

Hence 1 kg ball can have more K.E than 100 kg ball by doubling velocity.

4 0

3 years ago

A force of 20N pushes an object of mass 5.0kg along a rough surface of 5.0N

Alika [10]

Answer:

I'm sorry, I don't think there is any answer to give seeing as no question has been asked

7 0

3 years ago

Read 2 more answers

B. The role of the moon is greater than that of the sun in the occurrence of tides. ???​

B. The role of the moon is greater than that of the sun in the occurrence of tides. ???